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Re: Prize Puzzle : Primality Conjecture

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  • Adam
    When x=9, k=10, then A(9)=361 is not prime, T(10)=361 is a square, 10=k x=9, 9=x 3, T(k) = 361 is between 1 and B(x)=1681. Please contact me off list for
    Message 1 of 5 , Sep 10, 2008
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      When x=9, k=10, then A(9)=361 is not prime, T(10)=361 is a square,
      10=k > x=9, 9=x >3, T(k) = 361 is between 1 and B(x)=1681.

      Please contact me off list for mailing address for my $50 prize.

      There are 644 other counterexamples for x<=1000.

      Adam

      --- In primenumbers@yahoogroups.com, "aldrich617" <aldrich617@...>
      wrote:
      >
      > I offer a $50 prize to the first person who can submit
      > a verifiable counterexample or proof by 10/1/8
      > for the following primality conjecture:
      >
      > x, A(x), B(x), k, T(k) : integers;
      >
      > Let A(x) = 5x^2 - 5x +1;
      > Let B(x) = 25x^2 - 40x + 16;
      > Let T(k) = 5*(2*k -1)^2 - 4*A(x) ;
      >
      > If x > 3 and k > x , then A(x) will be prime if no value
      > T(k) exists such that 1 < T(k) < B(x) is a square.
      >
      > Aldrich Stevens
      >
    • Bernardo Boncompagni
      ... That s his point. There must be NO square in the selected interval, but T(10) is a square. By the way, this is just a restatement of Euler s factorization
      Message 2 of 5 , Sep 10, 2008
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        Adam dice:

        > When x=9, k=10, then A(9)=361 is not prime, T(10)=361 is a square,
        > 10=k > x=9, 9=x >3, T(k) = 361 is between 1 and B(x)=1681.

        That's his point. There must be NO square in the selected interval, but
        T(10) is a square. By the way, this is just a restatement of Euler's
        factorization method. If you put N = A(x) = m*n = (a-b)*(a+b) = a^2-b^2
        and do some algebraic tricks, it wouldn't be too difficult to prove the
        conjecture and claim your 50 bucks.

        Bernardo Boncompagni

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      • Jack Brennen
        You haven t given a counterexample. Read his conjecture again. Hint... his conjecture says nothing about cases where a square value DOES exist.
        Message 3 of 5 , Sep 10, 2008
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          You haven't given a counterexample. Read his conjecture again.

          Hint... his conjecture says nothing about cases where a square
          value DOES exist.

          Adam wrote:
          > When x=9, k=10, then A(9)=361 is not prime, T(10)=361 is a square,
          > 10=k > x=9, 9=x >3, T(k) = 361 is between 1 and B(x)=1681.
          >
          > Please contact me off list for mailing address for my $50 prize.
          >
          > There are 644 other counterexamples for x<=1000.
          >
          > Adam
          >
          > --- In primenumbers@yahoogroups.com, "aldrich617" <aldrich617@...>
          > wrote:
          >> I offer a $50 prize to the first person who can submit
          >> a verifiable counterexample or proof by 10/1/8
          >> for the following primality conjecture:
          >>
          >> x, A(x), B(x), k, T(k) : integers;
          >>
          >> Let A(x) = 5x^2 - 5x +1;
          >> Let B(x) = 25x^2 - 40x + 16;
          >> Let T(k) = 5*(2*k -1)^2 - 4*A(x) ;
          >>
          >> If x > 3 and k > x , then A(x) will be prime if no value
          >> T(k) exists such that 1 < T(k) < B(x) is a square.
          >>
          >> Aldrich Stevens
          >>
          >
          >
          >
          > ------------------------------------
          >
          > Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
          > The Prime Pages : http://www.primepages.org/
          >
          > Yahoo! Groups Links
          >
          >
          >
          >
          >
        • Peter Kosinar
          Hi Adam, ... Umm... pardon me? How does this contradict Aldrich s conjecture? As far as I understood the claim, A(x) is claimed to -be a prime- under the
          Message 4 of 5 , Sep 10, 2008
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            Hi Adam,

            Aldrich's conjecture:
            >> If x > 3 and k > x , then A(x) will be prime if no value
            >> T(k) exists such that 1 < T(k) < B(x) is a square.

            Adam's proposed counterexample:
            > When x=9, k=10, then A(9)=361 is not prime, T(10)=361 is a square,
            > 10=k > x=9, 9=x >3, T(k) = 361 is between 1 and B(x)=1681.

            Umm... pardon me? How does this contradict Aldrich's conjecture? As far as
            I understood the claim, A(x) is claimed to -be a prime- under the
            assumption that -no- value of T(k) between 1 and B(x) is a square.

            Peter

            --
            [Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278
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